Answer:
611.064 kJ
Explanation:
Given :
m = 200 mL = 200 g
Specific heat of ice = 2.06 J/g°C
Q = mcΔt
Δt = 0 - (-22) = 22
Q1 = 200 * 2.06 * 22 = 9064 J
Q2 = Melt 0 °C solid ice into 0 °C liquid water:
Q2 = m · ΔHf ; ΔHf = heat of fusion of water = 334j/g
Q2 = 200 * 334 = 66800 J
Q3 : Heat to convert from 0°C to 100°C
Q3 = mcΔt ; c = 4.19 J/g°C ; Δt = (100 - 0) = 100
Q3 = 200 * 4.19 * 100 = 83800 J
Q4: heat required to boil water to steam
Q = m · ΔHv
Hv = heat of vaporization of water = 2257 J/g
Q4 = 200 * 2257 = 451400 J
Total Q = Q1 + Q2 + Q3 + Q4
Q = 9064 + 66800 + 83800 + 451400
Q = 611,064 Joules
Q = 611.064 kJ
Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
The Answer is:
O 3s
Hope you got it right.
The weight would be 62.92.
